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Algebra Difficulty 4.7 AIME Prove it Greece

Let x>1x > 1 a non integer number. Prove that
(x+{x}[x]−[x]x+{x})+(x+[x]{x}−{x}x+[x])>92, \left( \frac{x+\{x\}}{[x]} - \frac{[x]}{x+\{x\}} \right) + \left( \frac{x+[x]}{\{x\}} - \frac{\{x\}}{x+[x]} \right) > \frac{9}{2},
where [x][x] and {x}\{x\} represents the integer and the fractional part of xx.

Solution

We put [x]=a[x] = a, {x}=r\{x\} = r, where 0≤r<10 \le r < 1. Then the given inequality becomes
(a+2ra−aa+2r)+(2a+rr−r2a+r)>92⇔2(ra+ar)−(aa+2r+r2a+r)>52. \left( \frac{a+2r}{a} - \frac{a}{a+2r} \right) + \left( \frac{2a+r}{r} - \frac{r}{2a+r} \right) > \frac{9}{2} \\ \Leftrightarrow 2 \left( \frac{r}{a} + \frac{a}{r} \right) - \left( \frac{a}{a+2r} + \frac{r}{2a+r} \right) > \frac{5}{2}.
Since ra+ar≥2\frac{r}{a} + \frac{a}{r} \ge 2, it is enough to prove that
aa+2r+r2a+r<32⇔0<2a2+11ar+2r2⇔2(a+r)2+7ar>0, \frac{a}{a+2r} + \frac{r}{2a+r} < \frac{3}{2} \Leftrightarrow 0 < 2a^2 + 11ar + 2r^2 \Leftrightarrow 2(a+r)^2 + 7ar > 0,
which is valid.

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